A: To start off you can imagine the space of time to exist on a straight line. The two spies would meet if they arrive within 5 minutes of each other. If spy A arrives at time \(x\) and spy B arrives at time \(y\) then they would meet if \(|x-y| < 5\). The inequality \(|x-y| < 5\) can be broken down as two equations
$$ x-y \lt 5 \\
y - x \lt 5 $$

The area included within the zone PQRS represents an area where the two would meet. The area of the square region is \(55 \times 55 = 3025\). The area of the lower triangle is \(\frac{1}{2}\times 50 \times 50 = 1250\). Thus the sought probability is $$ P(\text{meet}) = 1 - \frac{1250\times 2}{3025} = 17.35\% $$ Some of the best books to learn th…

Q: In a certain market there exists two brands of cereal from two competing companies. Brand X and Y. It is known from historical dataThe probability that a customer using brand X would hop to brand Y in a given year is \(12\%\)The probability that a customer using brand Y would hop to brand X in a given year is \(10\%\)After spending some money on advertising brand X is known to have \(30\%\) market share. What would the market share of X be after 3 years?

A: Situations like the above arise often in the industry and are best modelled using Transition Matrices. To begin with, assume the customer can be modelled as being in a particular "state" of loyalty. From the given data, this can shown in a transition matrix \(T\) form as below
$$
T = \pmatrix{ 0.88 & 0.12 \\ 0.1 & 0.9}
$$
The current state of market is the current state of the system which in turn can be represented as a row m…

Q: Alice has a bag with two red balls and a blue ball. There is another bag with two blue balls and one red ball. A ball is chosen randomly from Alice's bag and placed in the other bag. Next a random ball is chosen from that other bag and put back in Alice's bag. What is the probability that the remainder of the balls in her bag are red ifAlice draws a ball from her bag notices it to be a red ball Alice draws a ball from her bag and notices it to be a blue ball Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)
A: If Alice draws a red ball, the only way the other two balls will be red balls is if in the first draw a blue ball was chosen and in the next draw a red ball was chosen. This happens with probability \(\frac{1}{3}\) and \(\frac{1}{2}\) respectively. This yields an overall probability of \(\frac{1}{3}\times\frac{1}{2}=\frac{1}{6}\)

If the ball Alice draws is blue, there are only two scenarios that could have caused this.A red ball was…

A: The number of cars driving by a street in a given interval of time is a great example of a Poisson process which follows a distribution given by
$$f(k,\lambda) = \frac{(\lambda t)^{k}e^{-(\lambda t)}}{k!}$$
where \(k\) is the number Poisson events occurring in an interval of time \(t\) where the average rate of such events is \(\lambda\) per unit time (say minutes). The probability that an interval of time \(t\) would pass by without any events (i.e. \(k=0\) ) is
$$ P(\text{no event}) = e^{-\lambda t} $$
This implies that the probability of an event happening after time \(t\) is
$$F(t) = 1 - e^{-\lambda t}$$
The p.d.f of waiting times between events is
$$P(t) = \lambda e^{-\lambda t}$$
This is known as the exponential distribution and is good one to model arrival times of a P…

Q: You have an array of items that you want to shuffle. You pick an element and then pick a random number uniformly between \([1,n]\) and swap it with the element following which you move on to the next element and repeat. Would this result in a fair shuffle?

A: Some people shuffle a deck of playing cards in the above manner. They tend to pick a small (random sized) slab of cards from under the deck, place it on top and do so several times over.

The above shuffle would not result in a fair shuffle. Here is why. Assume you have \(n\) items in the list. When you pick an element and swap it out with another element there are \(n\) positions you could swap it with, including itself in which case there is no swap. This implies that the first swap would result in \(n\) arrangements. The second swap follows the same algorithm and would result in the same number of arrangements, that is \(n\). This leads to a…

Q: You routinely travel between A & B and have two streets you could drive through to get to B from A. There is a busy street that cuts through both streets so there are signals at X & Y (see fig). You are certain to be stopped at X and Y. Over a period of N days you want to ascertain which of the two routes is a busier one however as you drive you are in no position to count the number of vehicles in front of you and behind you. Assume you have no other data that you can collect other than what you can while driving your car. How do you find the busier street over a period of N days?

A: A surprisingly simple solution exists to find this. Take half the journeys \(\frac{N}{2}\) through route AXB and the other half through AYB. Simply keep track of the number of times you are stopped at the signal but are right in front of the queue facing the signal. This is easy to measure and count. In the le…